11.3.5: Energy Flow, Work and Heat

Let us return to the magnetic dipole model as shown in Figure 11.1.

In this section we will consider interactions with only one of the two environments. In Chapter 12 we will consider use of both environments, which will allow the machine to be used as a heat engine or refrigerator.

Consider first the case that the system is isolated from its environment, as shown in Figure 11.1 (the vertical bars represent barriers to interaction). The system is in some state, and we do not necessarily know which one, although the probability distribution \(p_i\) can be obtained from the Principle of Maximum Entropy. A change in state generally requires a nonzero amount of energy, because the different states have different energies. We can always imagine a small enough change \(dH\) in \(H\) so that the magnetic field cannot supply or absorb the necessary energy to change state. Then we can imagine a succession of such changes in \(H\), none of which can change state, but when taken together constitute a large enough change in \(H\) to be noticeable. We conclude that changing \(H\) for an isolated system does not by itself change the state. Thus the probability distribution \(p_i\) is unchanged. Of course changing \(H\) by an amount \(dH\) does change the energy through the resulting change in \(E_i(H)\):

\(dE = \displaystyle \sum_{i} p_i dE_i(H) \tag{11.41}\)

This change is reversible: if the field is changed back, the energy could be recovered in electrical or magnetic or mechanical form (there is nowhere else for it to go in this model). Energy flow of this sort, that can be recovered in electrical, magnetic, or mechanical form (or some other forms) is referred to as work. If \(dE\) > 0 then we say that work is positive, in that it was done by the external source on the system; if \(dE\) < 0 then we say that work is negative, in that it was done on the external source by the system. Naturally, in energy-conversion devices it is important to know whether the work is positive or negative. In many cases simply running the machine backwards changes the sign of the work; this is not always true of the other form of energy transfer, discussed below.

Changes to a system caused by a change in one or more of its parameters, when it cannot interact with its environment, are known as adiabatic changes. Since the probability distribution is not changed by them, they produce no change in entropy of the system. This is a general principle: adiabatic changes do not change the probability distribution and therefore conserve entropy.

First-order changes to the quantities of interest were given above in the general case where \(E\) and the various \(E_i\) are changed. If the change is adiabatic, then \(dE\) is caused only by the changes \(dE_i \) and the general equations simplify to

\(\begin{align*}
d p_{i} &=0 \tag{11.42} \\
d E &=\sum_{i} p_{i} d E_{i}(H) \tag{11.43}\\
d S &=0 \tag{11.44}\\
d \alpha &=-E d \beta-\beta \sum_{i} p_{i} d E_{i}(H) \tag{11.45}\\
0 &=\left[\sum_{i} p_{i}\left(E_{i}(H)-E\right)^{2}\right] d \beta+\beta \sum_{i} p_{i}\left(E_{i}(H)-E\right) d E_{i}(H) \tag{11.46}
\end{align*}\)

If, as in our magnetic-dipole model, the energies of the states are proportional to \(H\) then these adiabatic formulas simplify further to

\(\begin{align*}
d p_{i} &=0 \tag{11.47} \\
d E &=\left(\frac{E}{H}\right) d H \tag{11.48}\\
d S &=0 \tag{11.49}\\
d \alpha &=0 \tag{11.50} \\
d \beta &=-\left(\frac{\beta}{H}\right) d H \tag{11.51}
\end{align*}\)

Next, consider the system no longer isolated, but instead interacting with its environment. The interaction model permits heat to flow between the system and the environment, and by convention we will say the heat is positive if energy flows into the system from the environment, and negative if the energy flows the other way. Energy can be transferred by heat and work at the same time. Work is represented by changes in the energy of the individual states \(dE_i\), and heat by changes in the probabilities \(p_i\). Thus the formula for \(dE\) above becomes

\(dE = \displaystyle \sum_{i}E_i(H) dp_i + \sum_{i} p_i dE_i(H) \tag{11.52}\)

where the first term is heat and the second term is work.